Newton’s Law of Cooling
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PHXI11:THERMAL PROPERTIES OF MATTER

366652 A hot body placed in the air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being \(k\) times the temperature difference from the surrounding. Starting from \(t=0\), find the time at which the temperature difference between body and surrounding is equal to half of the initial temperature difference.

1 \(\ln \,2/k\)
2 \(k\ln 2\)
3 \(k/\ln 2\)
4 None of these
PHXI11:THERMAL PROPERTIES OF MATTER

366653 A body takes 5 minutes to cool from \(100^\circ C\) to \(60^\circ C\) and takes time to cool from \(80^\circ C\) to \(40^\circ C\). If surrounding temperature is \(20^\circ C\) find \(t\) :

1 \(9 \mathrm{~min}\)
2 \(4 \mathrm{~min}\)
3 \(6 \mathrm{~min}\)
4 \(8 \mathrm{~min}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366654 A body cools in a surrounding which is at a constant temperature of \(\theta_{0}\). Assume that it obeys Newton's law of cooling. Its temperature \(\theta\) is plotted against time \(t\). Tangents are drawn to the curve at the points \(P\left(\theta=\theta_{1}\right)\) and \(Q\left(\theta=\theta_{2}\right)\). These tangents meet the time axis at angles of \(\phi_{2}\) and \(\phi_{1}\), as shown
supporting img

1 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}\)
2 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}\)
3 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{1}}{\theta_{2}}\)
4 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{2}}{\theta_{1}}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366655 Consider two hot bodies \(B_{1}\) and \(B_{2}\) which have temperatures \(100^\circ C\) and \(80^\circ C\) respectively at \(t = 0\). The temperature of the surroundings is \(40^\circ C\). The ratio of the respective rates of cooling and of these two bodies at \(t = 0\) will be

1 \(R_{1}: R_{2}=5: 4\)
2 \(R_{1}: R_{2}=3: 2\)
3 \(R_{1}: R_{2}=4: 5\)
4 \(R_{1}: R_{2}=2: 3\)
PHXI11:THERMAL PROPERTIES OF MATTER

366656 A body cools in 7 minutes from \(60^\circ C\) to \(40^\circ C\). The temperature of the surrounding is \(10^\circ C\). The temperature of the body after the next 7 minutes will be

1 \(30^\circ C\)
2 \(34^\circ C\)
3 \(28^\circ C\)
4 \(32^\circ C\)
JEE - 2023
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PHXI11:THERMAL PROPERTIES OF MATTER

366652 A hot body placed in the air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being \(k\) times the temperature difference from the surrounding. Starting from \(t=0\), find the time at which the temperature difference between body and surrounding is equal to half of the initial temperature difference.

1 \(\ln \,2/k\)
2 \(k\ln 2\)
3 \(k/\ln 2\)
4 None of these
PHXI11:THERMAL PROPERTIES OF MATTER

366653 A body takes 5 minutes to cool from \(100^\circ C\) to \(60^\circ C\) and takes time to cool from \(80^\circ C\) to \(40^\circ C\). If surrounding temperature is \(20^\circ C\) find \(t\) :

1 \(9 \mathrm{~min}\)
2 \(4 \mathrm{~min}\)
3 \(6 \mathrm{~min}\)
4 \(8 \mathrm{~min}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366654 A body cools in a surrounding which is at a constant temperature of \(\theta_{0}\). Assume that it obeys Newton's law of cooling. Its temperature \(\theta\) is plotted against time \(t\). Tangents are drawn to the curve at the points \(P\left(\theta=\theta_{1}\right)\) and \(Q\left(\theta=\theta_{2}\right)\). These tangents meet the time axis at angles of \(\phi_{2}\) and \(\phi_{1}\), as shown
supporting img

1 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}\)
2 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}\)
3 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{1}}{\theta_{2}}\)
4 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{2}}{\theta_{1}}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366655 Consider two hot bodies \(B_{1}\) and \(B_{2}\) which have temperatures \(100^\circ C\) and \(80^\circ C\) respectively at \(t = 0\). The temperature of the surroundings is \(40^\circ C\). The ratio of the respective rates of cooling and of these two bodies at \(t = 0\) will be

1 \(R_{1}: R_{2}=5: 4\)
2 \(R_{1}: R_{2}=3: 2\)
3 \(R_{1}: R_{2}=4: 5\)
4 \(R_{1}: R_{2}=2: 3\)
PHXI11:THERMAL PROPERTIES OF MATTER

366656 A body cools in 7 minutes from \(60^\circ C\) to \(40^\circ C\). The temperature of the surrounding is \(10^\circ C\). The temperature of the body after the next 7 minutes will be

1 \(30^\circ C\)
2 \(34^\circ C\)
3 \(28^\circ C\)
4 \(32^\circ C\)
JEE - 2023
For More Updates join with Us
PHXI11:THERMAL PROPERTIES OF MATTER

366652 A hot body placed in the air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being \(k\) times the temperature difference from the surrounding. Starting from \(t=0\), find the time at which the temperature difference between body and surrounding is equal to half of the initial temperature difference.

1 \(\ln \,2/k\)
2 \(k\ln 2\)
3 \(k/\ln 2\)
4 None of these
PHXI11:THERMAL PROPERTIES OF MATTER

366653 A body takes 5 minutes to cool from \(100^\circ C\) to \(60^\circ C\) and takes time to cool from \(80^\circ C\) to \(40^\circ C\). If surrounding temperature is \(20^\circ C\) find \(t\) :

1 \(9 \mathrm{~min}\)
2 \(4 \mathrm{~min}\)
3 \(6 \mathrm{~min}\)
4 \(8 \mathrm{~min}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366654 A body cools in a surrounding which is at a constant temperature of \(\theta_{0}\). Assume that it obeys Newton's law of cooling. Its temperature \(\theta\) is plotted against time \(t\). Tangents are drawn to the curve at the points \(P\left(\theta=\theta_{1}\right)\) and \(Q\left(\theta=\theta_{2}\right)\). These tangents meet the time axis at angles of \(\phi_{2}\) and \(\phi_{1}\), as shown
supporting img

1 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}\)
2 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}\)
3 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{1}}{\theta_{2}}\)
4 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{2}}{\theta_{1}}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366655 Consider two hot bodies \(B_{1}\) and \(B_{2}\) which have temperatures \(100^\circ C\) and \(80^\circ C\) respectively at \(t = 0\). The temperature of the surroundings is \(40^\circ C\). The ratio of the respective rates of cooling and of these two bodies at \(t = 0\) will be

1 \(R_{1}: R_{2}=5: 4\)
2 \(R_{1}: R_{2}=3: 2\)
3 \(R_{1}: R_{2}=4: 5\)
4 \(R_{1}: R_{2}=2: 3\)
PHXI11:THERMAL PROPERTIES OF MATTER

366656 A body cools in 7 minutes from \(60^\circ C\) to \(40^\circ C\). The temperature of the surrounding is \(10^\circ C\). The temperature of the body after the next 7 minutes will be

1 \(30^\circ C\)
2 \(34^\circ C\)
3 \(28^\circ C\)
4 \(32^\circ C\)
JEE - 2023
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PHXI11:THERMAL PROPERTIES OF MATTER

366652 A hot body placed in the air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being \(k\) times the temperature difference from the surrounding. Starting from \(t=0\), find the time at which the temperature difference between body and surrounding is equal to half of the initial temperature difference.

1 \(\ln \,2/k\)
2 \(k\ln 2\)
3 \(k/\ln 2\)
4 None of these
PHXI11:THERMAL PROPERTIES OF MATTER

366653 A body takes 5 minutes to cool from \(100^\circ C\) to \(60^\circ C\) and takes time to cool from \(80^\circ C\) to \(40^\circ C\). If surrounding temperature is \(20^\circ C\) find \(t\) :

1 \(9 \mathrm{~min}\)
2 \(4 \mathrm{~min}\)
3 \(6 \mathrm{~min}\)
4 \(8 \mathrm{~min}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366654 A body cools in a surrounding which is at a constant temperature of \(\theta_{0}\). Assume that it obeys Newton's law of cooling. Its temperature \(\theta\) is plotted against time \(t\). Tangents are drawn to the curve at the points \(P\left(\theta=\theta_{1}\right)\) and \(Q\left(\theta=\theta_{2}\right)\). These tangents meet the time axis at angles of \(\phi_{2}\) and \(\phi_{1}\), as shown
supporting img

1 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}\)
2 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}\)
3 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{1}}{\theta_{2}}\)
4 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{2}}{\theta_{1}}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366655 Consider two hot bodies \(B_{1}\) and \(B_{2}\) which have temperatures \(100^\circ C\) and \(80^\circ C\) respectively at \(t = 0\). The temperature of the surroundings is \(40^\circ C\). The ratio of the respective rates of cooling and of these two bodies at \(t = 0\) will be

1 \(R_{1}: R_{2}=5: 4\)
2 \(R_{1}: R_{2}=3: 2\)
3 \(R_{1}: R_{2}=4: 5\)
4 \(R_{1}: R_{2}=2: 3\)
PHXI11:THERMAL PROPERTIES OF MATTER

366656 A body cools in 7 minutes from \(60^\circ C\) to \(40^\circ C\). The temperature of the surrounding is \(10^\circ C\). The temperature of the body after the next 7 minutes will be

1 \(30^\circ C\)
2 \(34^\circ C\)
3 \(28^\circ C\)
4 \(32^\circ C\)
JEE - 2023
Join With Us for More Updates
PHXI11:THERMAL PROPERTIES OF MATTER

366652 A hot body placed in the air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being \(k\) times the temperature difference from the surrounding. Starting from \(t=0\), find the time at which the temperature difference between body and surrounding is equal to half of the initial temperature difference.

1 \(\ln \,2/k\)
2 \(k\ln 2\)
3 \(k/\ln 2\)
4 None of these
PHXI11:THERMAL PROPERTIES OF MATTER

366653 A body takes 5 minutes to cool from \(100^\circ C\) to \(60^\circ C\) and takes time to cool from \(80^\circ C\) to \(40^\circ C\). If surrounding temperature is \(20^\circ C\) find \(t\) :

1 \(9 \mathrm{~min}\)
2 \(4 \mathrm{~min}\)
3 \(6 \mathrm{~min}\)
4 \(8 \mathrm{~min}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366654 A body cools in a surrounding which is at a constant temperature of \(\theta_{0}\). Assume that it obeys Newton's law of cooling. Its temperature \(\theta\) is plotted against time \(t\). Tangents are drawn to the curve at the points \(P\left(\theta=\theta_{1}\right)\) and \(Q\left(\theta=\theta_{2}\right)\). These tangents meet the time axis at angles of \(\phi_{2}\) and \(\phi_{1}\), as shown
supporting img

1 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}\)
2 \(\dfrac{\tan \phi_{2}}{\tan \phi_{1}}=\dfrac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}\)
3 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{1}}{\theta_{2}}\)
4 \(\dfrac{\tan \phi_{1}}{\tan \phi_{2}}=\dfrac{\theta_{2}}{\theta_{1}}\)
PHXI11:THERMAL PROPERTIES OF MATTER

366655 Consider two hot bodies \(B_{1}\) and \(B_{2}\) which have temperatures \(100^\circ C\) and \(80^\circ C\) respectively at \(t = 0\). The temperature of the surroundings is \(40^\circ C\). The ratio of the respective rates of cooling and of these two bodies at \(t = 0\) will be

1 \(R_{1}: R_{2}=5: 4\)
2 \(R_{1}: R_{2}=3: 2\)
3 \(R_{1}: R_{2}=4: 5\)
4 \(R_{1}: R_{2}=2: 3\)
PHXI11:THERMAL PROPERTIES OF MATTER

366656 A body cools in 7 minutes from \(60^\circ C\) to \(40^\circ C\). The temperature of the surrounding is \(10^\circ C\). The temperature of the body after the next 7 minutes will be

1 \(30^\circ C\)
2 \(34^\circ C\)
3 \(28^\circ C\)
4 \(32^\circ C\)
JEE - 2023